Complement of a compact set in the Riemann sphere has countable many components.

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I am reading Walter Rudin's "Real and Complex Analysis", and he states that given $P \subset \mathbb{C}$ a compact set, and $\hat{\mathbb{C}}$ the Riemann sphere, the complement of $P$ in $\hat{\mathbb{C}}$ (i.e. $ \hat{\mathbb{C}} -P$) has at most countable many components.

How can I prove this? It probably is pretty trivial since Rudin skips over the proof of the statement but I cannot seem to prove it.

I have tried covering $\hat{\mathbb{C}}$ with the connected components and a neighborhood of $P$. Then supposing you have an uncountable amount of components, tried get to a contradiction using the compactness of $\hat{\mathbb{C}}$.

But, how can I find a neighborhood of $P$ such that it does not cover an infinite amount of components? I am thinking in cases like when $P$ is a Cantor Space.

Because the Cantor Set in $\mathbb{R}$ is compact and its complement in the real line has uncountable many connected components.

So, it leaves me thinking what if $P$ behaves in a way that given any neighborhood $V$ of $P$, then $V$ covers an uncountable amount of connected components of $\hat{\mathbb{C}}-P$? Cannot think of any case that does, but cannot also negate that there is not such a case.

Any tips would be nice, Thanks in advanced.

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Take one point in the complement of your compact set off the sphere and identify the punctured sphere with $\mathbb{R}^2$ via a homeomorphism. There you can say that the complement of a compact set is open, and each connected component therefore contains at least one point with rational coordinates. Therefore there is an injection from the connected-components of your complement to $\mathbb{Q}^2$ which is countable.

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As manifolds are locally connected, components of open sets ( like the complement of compact sets) are open. And disjoint families of open sets in a separable space are at most countable.